Leetcode1252. Cells with Odd Values in a Matrix

public int oddCells(int n, int m, int[][] indices) {
    int[][] Array = new int[n][m];
    int rowlength = indices.length; //求出indices数组行数
    for (int i = 0; i < rowlength; i++) { //开始分别对每个索引的行列进行增加
        for (int j = 0; j < m; j++) {//先对行加
        Array[indices[i][0]][j]+=1;
        }
        for(int k=0;k<n;k++){//再对列加
            Array[k][indices[i][1]]+=1;
        }
    }
    //统计里面偶数个数
    int odd=0;
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++)
            if (Array[i][j]%2==1)
                odd++;
    }
    System.out.println(odd);
    return odd;
}
方法时间复杂度极低

 Solutions:

后来发现规律

其实根本不用求出矩阵中的元素

只需要根据行变换和列变换次数，既可以得出奇数和偶数

public int oddCells(int n, int m, int[][] indices) {
    int[] row = new int[n];
    int[] col = new int[m];
    for (int i = 0; i < indices.length; i++) {
        row[indices[i][0]] += 1;
        col[indices[i][1]] += 1;
    }
    int odd = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) {
            if ((row[i] + col[j]) % 2 == 1)
                odd++;
        }
    System.out.println(odd);
    return odd;
}

Leetcode1252. Cells with Odd Values in a Matrix

原文：https://www.cnblogs.com/chengxian/p/12204030.html